How do you Calculate the Δx for each of the following?
a.) an electron with Δv= 0.100 m/s
b.) a baseball with mass 145 g with Δv = 0.100 m/s
a.) an electron with Δv= 0.100 m/s
b.) a baseball with mass 145 g with Δv = 0.100 m/s
1 Answer
The version of the Heisenberg Uncertainty Principle as defined with uncertainties
\mathbf(DeltavecxDeltavecp_x >= h)
Recall from physics that momentum is
So, if you want to calculate the uncertainty in position:
DeltavecxmDeltavecv_x >= h
color(green)(Deltavecx >= h/(mDeltavecv_x))
You could resolve the inequality by simply solving for the minimum uncertainty of the x position. Then you would get:
-
color(blue)(Deltavecx) = h/(m_eDeltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/((9.109xx10^(-31) "kg")("0.100 m/s")) = color(blue)(7.27xx10^(-3) "m") -
color(blue)(Deltavecx) = h/(m_"ball"Deltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/(("0.145 kg")("0.100 m/s")) = color(blue)(4.57xx10^(-32) "m")
The physical interpretation is...
- You know to an OK certainty as to what the position of the baseball is. The radius of an average baseball is about
7.4xx10^(-2) "m" , so your uncertainty is less thanpm 10% . - For an electron, the uncertainty in velocity is quite low (an electron normally moves very quickly, and
"0.100 m/s" is sluggish for an electron), and thus, the uncertainty for its position is quite high.
Think about the radius of an electron, which is somewhere around